Quantum Physics - jjphysics [licensed for non …jjphysics.pbworks.com/f/10_11_H1_Quantum+Physics...A quantum of electromagnetic radiation is known as a photon. Eg.1: (Ans: E) Photon

8866 H1 Physics – J2/2011 11.Quantum Physics

Quantum Physics Content 1. Energy of a photon 2. The Photoelectric effect 3. Wave-particle duality 4. Energy levels in atoms 5. Line spectra

(a) Show appreciation of the particulate nature of electromagnetic radiation. (b) Recall and use E = hf. (c) Show an understanding that the photoelectric effect provides evidence for a

particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature.

(d) Recall the significance of threshold frequency. (e) Recall and use the equation 21

max2 mv = eVs, where Vs is the stopping potential. (f) Explain photoelectric phenomena in terms of photon energy and work function

energy. (g) Explain why the maximum photoelectric energy is independent of intensity whereas

the photoelectric current is proportional to intensity. (h) Recall, use and explain the significance of hf = φ + 21

max2 mv . (i) Describe and interpret qualitatively the evidence provided by electron diffraction for

the wave nature of particles.

(j) Recall and use the relation for the de Broglie wavelength λ = hp

.

(k) Show an understanding of the existence of discrete electron energy levels in isolated atoms (e.g. atomic hydrogen) and deduce how this leads to spectral lines.

(l) Distinguish between emission and absorption line spectra. (m) Recall and solve problems using the relation hf = E1 − E2.

jj_sph2011 Page 1 of 19 Sections marked * are extras


(a) Show an appreciation of the particulate nature of electromagnetic radiation

Electromagnetic energy is emitted, transmitted and absorbed in discrete packets or quanta. A quantum of electromagnetic radiation is known as a photon.

Eg.1: (Ans: E) Photon is the name given to A an electron emitted from a metal surface by the action of light. B a unit of energy. C a positively charged atomic particle. D an electron emitted from a metal surface by the action of heat. E a quantum of electromagnetic radiation. (b) Recall and use E = hf

The energy of a photon, E, is proportional to its frequency f, and is expressed as

E = hf = h cλ

where h is the Planck constant (= 6.63×10-34 Js).

Eg.2: Find the energy of a photon of electromagnetic radiation of wavelength 0.15 nm. In

which region of the electromagnetic spectrum is this radiation? Answer:

E = hf = hλc = (6.63×10-34 Js) −

××

8 -1

9

(3 10 ms )(0.15 10 m)

= 1.33×10-15 J.

X-ray. Recall:



Eg.3: (Ans: D) The wavelength of a 5 MeV γ-ray is A 4.95×10-38 m B 8.89×10-32 m C 8.89×10-30 m D 2.49×10-13 m E 2.48×10-10 m Answer: Data: 1 MeV = (1.6×10-19) x (1000 000) = 1.6×10-13 J

E = 5 MeV = 5 (1.6×10-13 J) = 8.0×10-13 J

E = hf = hλc λ = hc

E=

−

−

× ××

34 8 -1

13

(6.6 10 Js)(3 10 ms )(8.0 10 J)

= 2.49×10-13 m.

Eg.4: (Ans: D) A photon of light enters a block of glass after traveling through a vacuum. The

energy of the photon on entering the glass block A increases because its associated wavelength decreases. B decreases because the speed of the radiation decreases. C stays the same because the speed of the radiation and the associated

wavelength do not change. D stays the same because the frequency of the radiation does not change. E stays the same because the speed of the radiation and its wavelength

increase by the same order. Answer: Only speed and wavelength change, frequency is unchanged. (d) Recall the significance of threshold frequency (f) Explain photoelectric phenomena in terms of photon energy and work function

energy. (g) Explain why the maximum photoelectric energy is independent of intensity

whereas the photoelectric current is proportional to intensity

Photoelectric phenomena: Generally, electrons may be emitted from a metal surface by 3 methods: (1) by friction, e.g. by rubbing a metal surface vigorously with a piece of cloth, (2) by thermionic emission, i.e. by heating a metal such that the electrons in it

would gain enough kinetic energy to overcome the attraction of the metal surface and escape from the surface (almost as though they are boiled off the surface).



(3) by photoelectric effect, i.e. to incident a beam of light (em radiation) on a

metal surface such that the electrons in it would gain enough kinetic energy from the incident light to overcome the attraction of the metal surface and escape from the surface.

Photoelectric effect is the emission of electrons from the metal surface when an electromagnetic radiation of high enough frequency is incident on the metal.

Experimental Setup:

incident radiation mA C A V

Procedure:

In the above setup, radiation (or light) of known frequency f and intensity I is incident onto an emitting electrode (cathode C) placed within an evacuated glass envelope, together with the collecting electrode (anode A ). Electrons may be emitted by C and collected by A. The electric potential of A may be varied and made positive or negative (higher or lower) with respect to C. Refer to Annex A for results and explanation.

(c) Show an understanding that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature

It is well-known that only waves are able to exhibit phenomena such as interference and diffraction. Light as an example of an em radiation does exhibit such phenomena, and hence may be considered as a wave. On the other hand, the results of the photoelectric effect showed the inadequacy of the wave nature of light. By considering light as a wave, only the first result can be explained (i.e. photoelectric current ∝ intensity). All the experimental results can be explained adequately by considering light as a stream of photons. A photon is a quantum of em radiation, which may be considered as a particle without mass, just a packet of energy.

* [The explanation for the experimental results of the photoelectric effect was proposed

by none other than Albert Einstein. He was most famous for his Theory of Relativity, but he won his Nobel Prize for Physics due to this theory (photoelectric effect) which assumed that light may be considered as a stream of photons. The equation that summarises the photoelectric effect was named Einstein’s equation. ]



Eg.5: (Ans: B) Which one of the following statements, referring to photoelectric emission, is always

true? A No emission of electrons occurs for very low intensity. Not true. As long as there are photons with enough energy, there will

be electrons emitted. It is independent of intensity. B For a given metal there is a minimum frequency of radiation below which no

emission occurs. True. This is the threshold frequency for that metal. C The velocity of the emitted electrons is proportional to the intensity of the

incident radiation. Not true. It is the photoelectric current that is proportional to intensity. D The number of electrons emitted per second is independent of the intensity of

the incident radiation. Not true. Number of electrons emitted per second is a measure of the

photoelectric current, which is proportional to intensity. E The number of electrons emitted per second is proportional to the frequency

of the incident radiation. Not true. Number of electrons emitted per second is a measure of the

photoelectric current, which is proportional to intensity, not frequency. Eg.6: (Ans: E) A source emits monochromatic light of wavelength λ at power P. Given that h is the

Planck constant and c the speed of light, the rate of emission of photons is

A λ

Pch

B λcPh

C λ

hcP

D λ

Phc

E λPhc

Answer: Each photon has energy, E = hf. If N is the number of photons emitted in time t, the

total energy emitted in time t can be expressed as NE = Nhf = Nh λc

The source emits light at a power, P = total energy emitted time taken

= Nhft

=λ

Nhct

So, the rate of emission of photons, Nt

= λPhc

Eg.7: (Ans: A) A beam of monochromatic radiation falls on to a metal X and photoelectrons are

emitted. The rate of emission of photoelectrons will be double if A a beam of double the intensity is used. B radiation of double the frequency is used. C radiation of double the wavelength is used. D the thermodynamic temperature of the metal is doubled. E a metal with a work function half that of X is substituted for X.

Answer: Double the rate of emission of photoelectrons means double the photoelectric

current. Since photoelectric current ∝ intensity, the intensity must also be doubled.



(h) Recall, use and explain the significance of hf = φ + 21max2 mv

The above equation is known as the Einstein’s photoelectric equation. When an electron absorbs energy hf from an incident photon, spends a minimum energy φ to be emitted from the metal surface, it will have a maximum kinetic energy,

21

max2 mv = hf − φ Note: (i) 21

max2 mv belongs to an electron, measured by the stopping potential, Vs, of the electrode A, which is made negative (lower potential) with respect to C. This represents the minimum negative potential required to stop even the most energetic electron from reaching the electrode A (indicated by the zero current in the milliammeter), therefore,

21max2 mv = eVs.

(ii) φ is a characteristic of the electrode C, (= hfo, where fo is the threshold

frequency). and hf is the energy of a photon.

(e) Recall and use the equation 21max2 mv = eVs, where eVs is the stopping potential

This just means, for the photoelectron:

loss of kinetic energy = gain in electric potential energy

Einstein’s equation can hence be written as:

eVs = hf − φ or eVs = hf − hfo

or Vs = −( ) ( ) oh hf fe e

or Vs =φ

−( )h fe e

Vs

slope = he

fo f

− φe

Note: (i) The graph of Vs versus f has a gradient of he

which is always

constant. (ii) Different material of C will give a different value of φ, and

hence the graph will have different y-intercept.



Eg.8: (Ans: E) The result of an experiment to investigate the energy of photoelectrons emitted from

a metallic surface is represented by the figure below. stopping potential 0 frequency The gradient of the graph depends on the A intensity of the incident radiation. B wavelength of the incident radiation. C work function of the irradiated surface. D pressure of residual gas in contact with the surface. E ratio of the Planck constant to the electronic charge. Eg.9: (Ans: B) Light quanta each of energy 3.5×10-19 J fall on the cathode of a photocell. The

current through the cell is just reduced to zero by applying a reverse voltage to make the cathode 0.25 V positive with respect to the anode. The minimum energy required to remove an electron from the cathode is

A 2.9×10-19 J B 3.1×10-19 J C 3.5×10-19 J D 3.9×10-19 J E 6.4×10-19 J Answer: Data: hf = 3.5×10-19 J, Vs = 0.25 V. Question asks for minimum energy required to remove an electron from the cathode,

i.e. the work function energy of the cathode, φ. Using eVs = hf − φ φ = hf − eVs = (3.5×10-19 J) − (1.6×10-19 C)(0.25 V) = 3.1×10-19 J Eg.10: (Ans: A) In a photoelectric experiment, electrons are ejected from metals X and Y by light of

frequency f. The potential difference V required to stop the electrons is measured for various frequencies.

If Y has a greater work function than X, which graph illustrates the expected results? V A V B V C V D V E X X Y X Y Y X Y Y X 0 f 0 f 0 f 0 f 0 f Answer:

First, the gradient of the graph cannot change (always = he

), so answers are A or D.

If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept.




(i) Describe and interpret qualitatively the evidence provided by electron diffraction for the wave nature of particles

When a beam of electron is passed through a thin film of crystal (e.g. graphite ), the dispersion pattern of the emergent electrons produced on a screen ( coated with fluorescent, (c) ) is observed to be similar to the diffraction pattern produced by a beam X-ray (a well-known electromagnetic wave, (b) ). This is known as the electron diffraction phenomenon.

This phenomenon provides evidence for the wave nature of particles like electrons.

(j) Recall and use the relation for the de Broglie wavelength λ =ph

de Broglie suggested that: for a particle of momentum p (= mv), which exhibits wave behaviour, it will have an associated wavelength λ, given by

λ =ph

Similarly, for electromagnetic radiation of wavelength λ, which exhibits particle behaviour, it will have an associated momentum p, given by

p =λh

This is known as the de Broglie principle.

Eg.11: (Ans: D) Light of frequency 5×1014 Hz consists of photons of momentum A 4.0×10-40 kg m s-1 B 3.7×10-36 kg m s-1 C 1.7×10-28 kg m s-1 D 1.1×10-27 kg m s-1 E 3.3×10-19 kg m s-1 Answer:

p =λh

=chf

= 1-8

1434

ms 100.3Hz) 105)(Js 1063.6(

××× −

= 1.1×10-27 kg m s-1.


Eg.12: (Ans: E) A beam of light of wavelength λ is totally reflected at normal incidence by a plane

mirror. The intensity of the light is such that photons hit the mirror at a rate n. Given that the Planck constant is h, the force exerted on the mirror by this beam is

A nhλ B λ

nh C 2 nhλ D hnλ2 E

λnh2

Answer:

Each photon has a momentum p =λh

When totally reflected at normal incidence by a plane mirror,

the change in momentum = 2λh

Since there are n photons hitting the mirror per unit time,

the rate of change of momentum, tpΔ = 2n

λh

By Newton’s 2nd law of motion, force, F =tpΔ = 2n

λh

(k) Show an understanding of the existence of discrete energy levels in isolated

atoms (e.g. atomic hydrogen) and deduce how this leads to spectral lines

• The electrons orbiting an atom can only occupy certain allowed orbits.

• In these allowed orbits, the total kinetic and

potential energies of all the electrons in the atom will give rise to the specific energy state of the atom e.g. E1, E2 and so on.

• In the normal state or ground state of an atom, its electrons will occupy the lowest allowed orbit so that the energy of the atom is at its lowest.

• Any change usually involves the movement

of the electrons to higher allowed orbits. For the different allowed orbits occupied, the atom will have different but specific energy states, or excited states.

• In this transition to a higher orbit, the energy

absorbed by the atom in excitation, the excitation energy, is equal to the difference between the energy levels of the two states.

• The highest excited state is when the

electron is at infinity (i.e. the atom is ionised), so the highest excitation energy is the ionisation energy.

E4

E3

E2

E1

Energy level diagram of an atom having various allowed states. The lowest energy state E1 is the ground state. All others are excited states. Note the bigger gaps between energy levels in the lower region (say, between E2 and E1) than those in the higher region (say, between E4 and E3).



• On the other hand, if an electron makes a transition down from a higher excited

state to a lower energy level, the excess energy will be emitted as a photon, whose energy corresponds to the difference between the initial and final energy levels (hf = E2 − E1).

• Photons emitted from such transitions, when allowed to pass through a narrow slit and a diffraction grating, will give rise to a set of line spectrum, unique to the element.

High Potential Difference

Diffraction Grating

Line spectrum

Low Pressure Gas

Eg.13: (Ans: C) The minimum energy to ionise an atom is the energy required to A add one electron to the atom. B excite the atom from its ground state to its first excited state. C remove one outermost electron from the atom. D remove one innermost electron from the atom. E remove all the electrons from the atom. Remark: This is the ionisation energy of the atom.



E2

E1

12 EEλ

hchf −==

(l) Distinguish between emission and absorption line spectra

Emission Line Spectra for Hydrogen

• A spectrum is produced by using a prism or diffraction grating to separate the

various wavelengths in a beam of light. It is the set of wavelengths that are observed.

• A line spectrum contains a discrete set of wavelengths, like the ones shown here. It is a characteristic feature of an element. Examples of emission line spectra are as seen above.

Emission

An excited electron transits to a lower energy

level by emitting a photon. • Emission refers to the emission of a specific amount of energy from an isolated

atom (e.g. an atom of an element in vapour or gaseous state, at low pressure) when its electrons jump from higher to lower allowed orbits, releasing the specific excess energy.

• Such specific energy E released can only take the form of a photon, with the

corresponding frequency f related by

E = hf = 12 EEλ

hc−=

• An emission line spectrum of an element consists of a series of separate bright lines of definite frequencies (or wavelengths) on a dark background. It is produced when a stream of photons of different frequencies is passed through a narrow slit and normally through a diffraction grating.

• These photons are emitted randomly from transitions (from higher to lower

excited states or the ground state) in the excited atoms of the element in a vapour or gas at low pressure.



• Since each element has a unique set of orbital electrons, the emission line

spectrum of an element is also unique, enabling it to be used as a means of identification of the element.

• Atoms in solid state (e.g. a tungsten filament) are very close to one another, no

longer isolated. The energy levels of each atom are no longer identical to one another. Emission from these atoms will result in a continuous spectrum (i.e. the numerous lines diffracted on the screen are so close to each other that separation between these lines are negligible and the spectrum appears to be continuous).

Eg.14: (Ans: B) Which of the following provides experimental evidence for discrete electron energy

levels in atoms? A the spectrum of a tungsten filament lamp. B the spectrum of a sodium discharge lamp. C the photoelectric effect. D the emission of β-particles by radioactive atoms. E the emission of γ-rays by radioactive atoms. Remark: Sodium in a discharge lamp is in gaseous state. Tungsten filament is a solid. Eg.15: (Ans: A) The existence of energy levels within atoms can be demonstrated directly by

observing that A atoms can emit spectra. B photoelectrons are only emitted for wavelength greater than a critical

wavelength. C some α-particles are reflected back through very large angles by atoms in a

solid. D X-rays with frequencies up to a certain maximum are emitted by a target. E atoms in a solid diffract electrons in the same way as crystals diffract X-rays.

Absorption

• Absorption refers to the absorption of energy by an atom from external sources. • Absorption can occur only if it results in the atom being excited to a higher

allowed energy state, i.e. only a specific amount of energy EΔ can be absorbed. • No absorption will occur if too little or too much energy is given to the atom.

E2

12 EEE −=Δ

E1

An electron transits to a higher energy level after absorption.



There are 3 ways in which an atom can absorb energy: (1) from thermal energy (e.g. when an element is heated by a bunsen flame); or (2) from kinetic energy of a colliding particle (e.g. when an atom in a vapour in a tube at low pressure is collided by an

electron moving across the tube, say, in an electric fluorescent lamp.); or (3) from an incident photon (e.g. when a beam of bright light is incident on a

vapour or gas.)

Note: For (1) and (2), an atom will absorb the exact amount of energy required for a particular transition, which may be any fraction of the energy supplied by the source. For (3), only photons with the exact amount of energy required for the transition will be absorbed. Photons with energies lower or higher than this exact amount will not be absorbed. In the diagram below, the bright light beam originates from the filament lamp. The sodium lamp does not give out light on its own, consisting of only sodium vapour. The resulting light beams would pass through the collimator which makes the beams parallel and incident normally onto the diffraction grating. The spectrum formed is observed using the telescope.

• An absorption line spectrum of an element consists of a series of separate dark lines of definite frequencies (or wavelengths) on a coloured background.

• The coloured background is produced when a stream of photons of different

frequencies from a white light source (e.g. tungsten filament lamp) is passed through a narrow slit and a diffraction grating.

• The cool vapour of the element concerned is placed between the white light

source and the narrow slit, such that its atoms may absorb excitation energy from photons incident from the white light source.

• The unabsorbed photons from the white light source will be incident on the

screen with the original intensity.

• After the absorption, the excited atoms will eventually go back to the ground state by emitting the same photons absorbed earlier.



• The re-emitted photons from the excited atoms in the cool vapour of the element will be in all directions. Only a small fraction of the re-emitted photons will be incident onto the screen, hence they will form lines with a much lower intensity.

• This accounts for the dark lines in the spectrum, which appear on the exact

positions as the corresponding emission line spectrum of the element.

• Note that these dark lines are not totally dark. In fact, they are of the same colour as their neighbouring background. They only appear dark against the coloured background because they are too dim, by comparison.

Summary:

Distinguishing between In terms of

Emission Line Spectra Absorption Line Spectra

Initial state of gas atoms

• produced by hot (gas atoms are initially at excited state) vapour or gas.

• produced when white light passes through cool (gas atoms are initially at ground state) vapour or gas.

Spectrum Pattern

• a series of separate bright lines of definite wavelength on a dark background

• a series of separate dark lines of definite wavelength on a coloured background

Explanation • Excited atoms emit photons

with the characteristic wavelengths corresponding to the transitions from higher to lower energy levels.

• Wavelengths corresponding to bright lines on the spectrum are characteristics of the element emitting the light.

• Cool gas absorbs photons with the characteristic wavelengths corresponding to the transitions from lower to higher energy levels.

• Subsequently, the excited atoms transit back to the lower energy levels by emitting the same photons in all directions Small fraction of emitted photons incident on screen lines of lower intensity formed (appear as dark lines on the spectrum)

• Wavelengths corresponding to dark lines on the spectrum are characteristics of the element emitting the light.



Eg.16: (Ans: E) When a parallel beam of white light passes through a metal vapour, dark lines

appear in the spectrum of the emergent light. This is principally because energy is absorbed and

A is not re-emitted at all. B is re-radiated as infra-red. C is re-radiated as ultra-violet. D is re-radiated gradually over a long period of time. E is re-radiated uniformly in all directions.

Eg.17: (Ans: C) White light from a tungsten filament lamp is passed through sodium vapour and

viewed through a diffraction grating. Which of the following best describes the spectrum which would be seen? A coloured lines on a black background B coloured lines on a white background C dark lines on a coloured background D dark lines on a white background Remark: The coloured background is due to the white light from tungsten lamp forming a

continuous spectrum. (m) Recall and solve problems using the relation hf = E1 − E2.

In a transition from an initial level of energy E1 to a final level of energy E2 ( E1 > E2 ), a photon of frequency f is emitted, with its energy E given by

E = hf = hλc = E1 − E2



Eg.18: (Ans: D) An atom emits a spectral line of wavelength λ when an electron makes a transition

between the levels E1 and E2. Which expression correctly relates λ, E1 and E2?

A λ =ch ( E1 − E2) B λ = ch ( E1 − E2)

C λ =)( 21 EEh

c−

D λ =21 EE

ch−

Answer:

hf = E1 − E2 hλc = E1 − E2 λ =

21 EEhc−

Eg.19: (Ans: A) In the figure below, E1 to E6 represent some of the energy levels of an electron in the

hydrogen atom. E6 −0.38 eV E5 −0.54 eV E4 −0.85 eV E3 −1.5 eV E2 −3.4 eV E1 −13.6 eV Which one of the following transitions produces a photon of wavelength in the ultra-

violet region of the electro-magnetic spectrum? [ 1 eV = 1.6×10-19 J. ] A E2 E1 B E3 E2 C E4 E3 D E5 E4 E E6 E5 Answer: Assume a transition from E2 to E1, the resulting photon emitted would have an

energy, E = (−3.4 eV) − (−13.6 eV) = 10.2 eV = 10.2(1.6×10-19 J) = 1.63×10-18 J. The frequency and wavelength of the photon are related by the equation

E = hf = hλc

λ = hEc = (6.63×10-34 Js)

J 1063.1ms 103

18

-18

−××

= 1.22×10-7 m This corresponds to the wavelength of ultra-violet radiation.




Eg.20: (Ans: E) The diagram below represents, drawn to scale, the energy levels for an electron in a

certain atom. energy E4 E3 E2 E1 The transition from E3 to E1 produces a green line. What transition could give rise to

a red line? A E4 to E3 B E4 to E2 C E4 to E1 D E3 to E2 E E2 to E1 Remark: • Frequency of red light is lower than frequency of green light, so energy of a

photon from a red light is lower than that from a green light. • But gaps between energy levels are bigger for lower levels than for higher levels. • The transition from E3 to E1 gives a green light (in the visible light region of the

em spectrum), then to give a red light, the transition must also be to E1. • Transition to other levels would give a photon of much lower energy, regardless

of where the transition is originated from.



Annex A: Results and explanations for the photoelectric effect

Explanation

Result Experimental Observation Considering light as a wave Considering light as a particle

A beam of light carries a continuous flow of energy when it is incident onto the metal surface. The intensity of the beam is proportional to the square of the wave amplitude. Intensity is a measure of the rate at which energy would be imparted onto the electrons per unit surface area of the emitting electrode C.

Since each photon has a specific amount of energy (= hf), then the intensity of a beam of light is a measure of the rate of incidence of photons, i.e. the number of photons incident per unit time. The higher the intensity, the more photons per unit time is incident onto the emitting electrode C.

1 The rate of emission of photoelectrons ( measured as photoelectric current ) is proportional to the incident radiation intensity. Intensity is the rate of incidence of energy per unit area.

The higher the intensity of the light beam, the higher the energy imparted onto the electrons. This resulted in a higher number of photoelectrons emitted per unit time which means a higher current. Can explain observation.

It is assumed each photon gives all its energy (= hf) to a single electron in the metal. Hence the more photons incident onto the metal surface, the more photoelectrons emitted. Can explain observation.

2 For every material of C irradiated, there is a minimum frequency ( threshold frequency fo ) required to liberate an electron from the surface. This fo is independent of the light intensity.

Emission of photoelectrons from the metal surface should occur at any frequency. It is independent of frequency. Cannot explain observation.

There is a certain minimum energy called the work function energy, φ necessary to liberate an electron from its surface. Since the energy of a photon is expressed as hf, hence the minimum photon energy required to liberate an electron from its surface must be hfo (= φ) where fo is the threshold frequency. fo andφ are characteristic of the material C. Can explain observation.

photoelectric current

Intensity



Explanation

Result Experimental Observation Considering light as a wave Considering light as a particle

3 The maximum kinetic ( photoelectric ) energy of emitted electrons depends only on the frequency of the incident radiation, and not its intensity.

The higher the intensity, the higher the energy imparted onto the electrons on the metal surface. The electrons should then be emitted with higher kinetic energies, which contradict the experimental result. Cannot explain observation.

The emitted photoelectrons would have absorbed energy from the incident photons. Since each electron absorbs energy only from one photon, what matters is the energy of each photon (which depends only on its frequency). High intensity means there is a large number of photons incident onto the electrode C per unit time, but this does not affect the energy absorbed by a single electron. Can explain observation.

current Frequency fixed, Intensity I2 > I1

I2

I1

V Vs current

Intensity fixed, Frequency f2 > f1

f2 f1

V VS1 VS2

4 The emission of photoelectrons starts with no observable time lag, even for very low intensity of incident radiation.

For very low intensity radiation, an electron would require some time to absorb enough energy to be emitted from the metal surface. Then there would be a certain time lag in its emission. Cannot explain observation.

For the incident photon, its energy will be instantaneously transferred to the absorbing electron, in a one-to-one interaction. This accounts for the instantaneous emission of the photoelectron. Can explain observation.

Documents

Quantum Physics - jjphysics [licensed for non …jjphysics.pbworks.com/f/10_11_H1_Quantum+Physics...A quantum of electromagnetic radiation is known as a photon. Eg.1: (Ans: E) Photon